The change in enthalpy for the forward reaction is -91KJ/mol
CO(g) + 2H2(g) <-> CH3OH(g)
1. The forward reaction is?
A. Endothermic
B. Exothermic **
2. If the reaction is at equilibrium and then was heated _____ CH3OH would be present after the reaction adjusted to the new temperature
A.more
B.less
C.same amount
3. If the reaction was at equilibrium and then the pressure in its container was increased_____CH3OH would be present after the reaction adjusted to the new pressure
A.more
B.less
C.same amount
4. If the reaction was at equilibrium and then H2 was added, _____ CH3OH would be present after the reaction adjusts
A.more
B.less
C.same amount
5. If reaction was at equilibrium and then H2 was added, ____CO would be present after the reaction adjusted
A.more
B.less
C.same amount
(I’m very confused and my teacher doesn’t help me)
______________
For question 1. I got the same answer.
I am confused because it says “after the reaction adjusts” so I thought when the reaction goes back to equilibrium it would all be the same amount again? I’m not sure I’m still confused.
Let me try question 3.
If the pressure increases then it would shift towards the side where less molecules are. Therefore it would shift to the right. I just don’t know how to tell if there will be more, less or same amount.
Since the pressure is upsetting the equilibrium and shifting to the right. There would be “more” CH3OH(g).?
Question 4. If H2 was added how much CH3OH?
If we add H2 it would shift to the right. Therefore they would be the same amount because it is taking H2 to make more CH3OH?
Question 5. If H2 was added.
If we add H2 it would shift to the right to try and make up that. So therefore, there would be less CO because it is getting taken to make up for CH3OH
The change in enthalpy for the forward reaction is -91KJ/mol
CO(g) + 2H2(g) <-> CH3OH(g)
1. The forward reaction is?
A. Endothermic
B. Exothermic **
You're right. delta H is negative means exothermic; delta H is positive means endothermic. Let me rewrite the equation to show that.
CO(g) + 2H2(g) <-> CH3OH(g) + heat
EXOTHERMIC so heat is given off so I've added + heat to show that.
2. If the reaction is at equilibrium and then was heated _____ CH3OH would be present after the reaction adjusted to the new temperature
A.more
B.less
C.same amount
Here is a succinct statement about Le Chatelier's Principle (which is what this question is all about). It won't win any Pulitizer Prizes but students understand it better than the formal textbook statement. "When a system in equilibrium is subjected to a stress it will shift in such a way as to undo what we did to it." So here is the equilibrium. CO(g) + 2H2(g) <-> CH3OH(g) + heat
So if we ADD HEAT the system will shift to use up the added heat. How can it do that? It can cause CH3OH to be heated and produce CO and H2. That is it will reverse the direction and go backwards (shift to the left) in order to use up the heat that was added. So you have CH3OH being used up (making it less) and it produces MORE CO and MORE H2 than was there when it was first equilibrated. Notice that the AMOUNT of CO is increased, the AMOUNT of H2 is increased and the AMOUNT of CH3OH is decreased.
3. If the reaction was at equilibrium and then the pressure in its container was increased_____CH3OH would be present after the reaction adjusted to the new pressure
See below
A.more
B.less
C.same amount
4. If the reaction was at equilibrium and then H2 was added, _____ CH3OH would be present after the reaction adjusts
A.more
B.less
C.same amount
5. If reaction was at equilibrium and then H2 was added, ____CO would be present after the reaction adjusted
A.more
B.less
C.same amount
(I’m very confused and my teacher doesn’t help me)
______________
For question 1. I got the same answer. I think you meant question 2. Question 1 was endothermic or exothermic.
I am confused because it says “after the reaction adjusts” so I thought when the reaction goes back to equilibrium it would all be the same amount again? I’m not sure I’m still confused. You're partly right. Yes it shifts to the left and produces less CH3OH and leaves more CO and H2 but OF COURSE it changes the amout as I discussed above.
Let me try question 3.
If the pressure increases then it would shift towards the side where less molecules are. Therefore it would shift to the right. I just don’t know how to tell if there will be more, less or same amount.
Since the pressure is upsetting the equilibrium and shifting to the right. There would be “more” CH3OH(g).?
Sure you do. You have the shift correct. If it shifts to the right you KNOW more CH3OH is produced and you KNOW CO and H2 are decreased. Yes, the amounts change. They MUST change. Why? Because Kc (or Kp) is a constant. A constant is a constant. In this case Kp = pCH3OH/pCO*p^2 H2. If you upset the equilibrium by changing the pressure and it shifts to the right, you KNOW the values for CO, H2 and CH3OH must change so that Kp, a constant, remains Kp, a constant. You can change pressure, volume, concentration and temperature. Changing T PROBABLY will change Kp/Kc but in all of the others Kp/Kd remains the original Kp and does not change. P and V will not cause the reaction to shift if moles on left = moles on the right.
Be sure and follow up if any of this is unclear. By the way, thank you so much for showing your confusion. It helped me answer because it immediately told me the problem you were having. I think you had Le Chatelier's Principle well in hand; however, you though the values for pressure or the values for cocentration or moles did not change. They do. Rmember, Kp or Kc is a constant and it reamins the same EXCEPT when changing the temperature. That will change Kp or Kc MOST of the time.
Question 4. If H2 was added how much CH3OH?
If we add H2 it would shift to the right. Therefore they would be the same amount because it is taking H2 to make more CH3OH?
Yes it will shift to the right because it wants use up the added H2 gas so now you will have more CH3OH and less H2.
Question 5. If H2 was added.
If we add H2 it would shift to the right to try and make up that. So therefore, there would be less CO because it is getting taken to make up for CH3OH
You're right. You add H2, the reaction shifts to the right, the amout of CH3OH increases and the amont of CO decreases.
1. B. Exothermic (Because the change in enthalpy is negative, indicating that heat is being released during the forward reaction)
2. B. less (When the reaction is heated, it will try to counteract the increase in temperature by shifting in the direction that absorbs heat, which is the reverse reaction. This will result in a decrease in the amount of CH3OH.)
3. A. more (Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas molecules. In this case, it will shift to the right, resulting in an increase in the amount of CH3OH.)
4. C. same amount (Adding more H2 will not affect the equilibrium position because H2 is not directly involved in the production of CH3OH. Therefore, the amount of CH3OH will remain the same.)
5. B. less (Adding more H2 will cause the equilibrium to shift towards the right to counteract the increase in H2. As a result, the amount of CO will decrease.)
Let's go through each question step by step:
1. The change in enthalpy for the forward reaction is -91 KJ/mol. Based on this information, we can determine the answer to be B. Exothermic. A negative change in enthalpy indicates an exothermic reaction, meaning heat is released.
2. If the reaction is at equilibrium and then heated, we need to consider Le Chatelier's principle. By heating the reaction, we are adding energy to the system. According to Le Chatelier's principle, the system will shift in the direction that opposes the change. In this case, the forward reaction is exothermic, so if we add heat, the system will shift to the left, decreasing the amount of CH3OH. The answer is B. Less CH3OH would be present after the reaction adjusted to the new temperature.
3. If the pressure in the container is increased, again we can apply Le Chatelier's principle. Increasing pressure will favor the side of the reaction with fewer moles of gas. In this case, the forward reaction has one molecule of CO and two molecules of H2, while the reverse reaction has one molecule of CH3OH. The forward reaction has fewer moles of gas, so when the pressure is increased, the system will shift to the right, towards the side with more moles of gas. Therefore, there will be more CH3OH present after the reaction adjusts. The answer is A. More CH3OH would be present.
4. If H2 is added to the system, Le Chatelier's principle can once again be applied. Adding more H2 will shift the reaction to the right, favoring the production of CH3OH. Since the reaction is at equilibrium, the amount of CH3OH will remain the same as the reaction adjusts. The answer is C. The same amount of CH3OH would be present.
5. Adding H2 will again shift the reaction to the right according to Le Chatelier's principle. However, this time, CO is consumed to produce more CH3OH. Therefore, there will be less CO present after the reaction adjusts. The answer is B. Less CO would be present.
I hope this helps clarify the answers to your questions!
To determine the answers to these questions, you need to consider the concept of Le Chatelier's principle. This principle states that if a system at equilibrium is disturbed by any changes such as temperature, pressure, or concentration, the system will adjust itself in such a way as to counteract the change and restore equilibrium.
Let's go through each question and explain the reasoning behind the answers:
1. The change in enthalpy for the forward reaction is -91 KJ/mol. Enthalpy is a measure of heat energy, and a negative value indicates an exothermic reaction. In an exothermic reaction, heat is released as a product, so the forward reaction is exothermic. Therefore, the correct answer is B. Exothermic.
2. If the reaction is at equilibrium and then heated, Le Chatelier's principle tells us that the system will shift in the direction that absorbs heat. Since the forward reaction is exothermic (heat is released), heating the system would be equivalent to removing heat. To restore equilibrium, the system would shift to the side that produces more heat, which is the reverse reaction. Consequently, the amount of CH3OH present after the reaction adjusts would be less. Therefore, the answer is B. Less.
3. If the pressure in the container is increased, the system will shift in the direction that reduces the pressure. According to the balanced equation, the forward reaction produces 1 molecule of CH3OH, while the reverse reaction involves more molecules (CO and 2H2). Therefore, increasing the pressure would favor the side with fewer molecules, which is the forward reaction. Consequently, the amount of CH3OH present after the reaction adjusts would be more. Thus, the answer is A. More.
4. If H2 is added to the system, this would increase the concentration of one of the reactants. Le Chatelier's principle tells us that the system will shift in the direction that reduces the concentration of the added substance. In this case, it means shifting towards the forward reaction. Therefore, there will be more CH3OH present after the reaction adjusts. Thus, the answer is A. More.
5. Similarly to question 4, if H2 is added, the system will shift towards the forward reaction. Since the forward reaction consumes CO to produce CH3OH, there would be less CO present after the reaction adjusts. Therefore, the answer is B. Less.
Remember, these explanations are based on the concept of Le Chatelier's principle and the understanding of how certain factors, such as temperature, pressure, and concentration, affect the equilibrium of a reaction.